javascript - Why does using parseInt on Error return 14? -
consider following:
parseint(new array(), 10); // -> nan parseint(new array(), 16); // -> nan parseint(new error(), 10); // -> nan parseint(new error(), 16); // -> 14
it seems behavior unique error/instances of error. can provide insight?
basically, that's because:
new error().tostring()
yields"error"
, andparseint("error", 16)
yields14
(because0xe
14
, parser stops @r
).
on other hand, new array()
not trigger same behavior because tostring()
method of array objects returns contents of array, delimited commas, not class name. therefore, new array().tostring()
yields empty string, , parseint()
subsequently yields nan
.
Comments
Post a Comment